Chapter 2, Problem 2.17P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Explanation of Solution

The given reaction is $2N{O}_2{}_{\left(g\right)}\to N_2{O}_4{}_{\left(g\right)}$.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Substituting the values,

$\begin{array}{l}\Delta H_{rxn}^o=\Delta H^0{}_{\left(N_2{O}_4\right)}-\left(2\times \Delta H^0{}_{\left(N{O}_2\right)}\right)\\ \\ \Delta H_{rxn}^o=\left(9.16kJmo{l}^{-1}\right)-\left(2\times 33.18kJmo{l}^{-1}\right)\\ \\ \Delta H_{rxn}^o=-57.29kJmo{l}^{-1}\end{array}$

The standard enthalpy change for the given reaction is $-57.29kJmo{l}^{-1}$

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Explanation of Solution

The given reaction is $N{O}_2{}_{\left(g\right)}\to 1/2N_2{O}_4{}_{\left(g\right)}$.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Substituting the values,

$\begin{array}{l}\Delta H_{rxn}^o=\left(1/2\Delta H^0{}_{\left(N_2{O}_4\right)}\right)-\left(\Delta H^0{}_{\left(N{O}_2\right)}\right)\\ \\ \Delta H_{rxn}^o=\left(1/2\times 9.16kJmo{l}^{-1}\right)-\left(\times 33.18kJmo{l}^{-1}\right)\\ \\ \Delta H_{rxn}^o=-28.6kJmo{l}^{-1}\end{array}$

The standard enthalpy change for the given reaction is $-28.6kJmo{l}^{-1}$

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Explanation of Solution

The given reaction is $3N{O}_2{}_{\left(g\right)}+H_{2}O\to 2HN{O}_3{}_{\left(aq\right)}+N{O}_{\left(g\right)}$.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Substituting the values,

$\begin{array}{l}\Delta H_{rxn}^o=\left(2\Delta H^0{}_{\left(HN{O}_3\right)}+\Delta H^0{}_{\left(NO\right)}\right)-\left(3\times \Delta H^0{}_{\left(N{O}_2\right)}+\Delta H^0{}_{\left(H_{2}O\right)}\right)\\ \\ \Delta H_{rxn}^o=\left(2\times -207.36kJmo{l}^{-1}+90.25kJmo{l}^{-1}\right)-\left(3\times 33.18kJmo{l}^{-1}+\left(-285.83kJmo{l}^{-1}\right)\right)\\ \\ \Delta H_{rxn}^o=-138.2kJmo{l}^{-1}\end{array}$

The standard enthalpy change for the given reaction is $-138.2kJmo{l}^{-1}$

(d)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Explanation of Solution

The given reaction is $cyclpropan{e}_{\left(g\right)}\to propen{e}_{\left(g\right)}$.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Substituting the values,

$\begin{array}{l}\Delta H_{rxn}^o=\left(2\Delta H^0{}_{\left(propene\right)}\right)-\left(\Delta H^0{}_{\left(cyclopropane\right)}\right)\\ \\ \Delta H_{rxn}^o=\left(+20.42kJmo{l}^{-1}\right)-\left(53.30kJmo{l}^{-1}\right)\\ \\ \Delta H_{rxn}^o=-32.88kJmo{l}^{-1}\end{array}$

The standard enthalpy change for the given reaction is $-32.88kJmo{l}^{-1}$

(e)

Interpretation Introduction

Interpretation:

The standard enthalpy for the given reaction has to be calculated.

Concept Introduction:

Bond dissociation enthalpy is an energy required to break one mole of gaseous bonds to form gaseous atoms.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Explanation of Solution

Net ionic equation is

$H^{+}{}_{\left(aq\right)}+C{l}^{-}{}_{{}_{\left(aq\right)}}+N{a}^{+}{}_{{}_{\left(aq\right)}}+O{H}^{-}{}_{{}_{\left(aq\right)}}\to C{l}^{-}{}_{{}_{\left(aq\right)}}+N{a}^{+}{}_{{}_{\left(aq\right)}}+H_2{O}_{{}_{{}_{\left(l\right)}}}$

The given reaction is $H^{+}{}_{\left(aq\right)}+O{H}^{-}{}_{{}_{\left(aq\right)}}\to H_2{O}_{{}_{{}_{\left(aq\right)}}}$.

Standard enthalpy change for the reaction can be calculated as,

$\Delta H_{rxn}^o=\underset{products}{\sum \Delta H_f^o}-\underset{\text{reactants}}{\sum \Delta H_f^o}$

Substituting the values,

$\begin{array}{l}\Delta H_{rxn}^o=\left(\Delta H^0{}_{\left(H_{2}O\right)}\right)-\left(\Delta H^0{}_{\left(H^{+}\right)}+\Delta H^0{}_{\left(O{H}^{-}\right)}\right)\\ \\ \Delta H_{rxn}^o=\left(-285.36kJmo{l}^{-1}\right)-\left(0kJmo{l}^{-1}+\left(-229.99kJmo{l}^{-1}\right)\right)\\ \\ \Delta H_{rxn}^o=-55.84kJmo{l}^{-1}\end{array}$

The standard enthalpy change for the given reaction is $-55.84kJmo{l}^{-1}$